Proving Bounded Subsets of R are Totally Bounded

[jdcasey9]

Homework Statement

Prove that a bounded subset of R is totally bounded.

Homework Equations

The Attempt at a Solution

Fix E > 0. Let A be subset of R, x be contained in A, and B(E/2, a) where E/2 is the radius of the ball and a is the center.

Assume that B(E/2, a) is closed (since a similar open ball is contained in the closed one we can infer it is true as well). We can find balls around each x (contained in A) that can be described by B(E, x) (open or closed does not matter).

For B(E/2, a), we can always find d(x, a) <= E/2 < E...

I am having trouble describing the obvious assumption that the union of B(E, x)'s will contain (and more) all of B(E/2), do I need to use the triangle inequality? Or is a written explanation reason enough?

 

[Dick]

Homework Statement

Prove that a bounded subset of R is totally bounded.

Homework Equations

The Attempt at a Solution

Fix E > 0. Let A be subset of R, x be contained in A, and B(E/2, a) where E/2 is the radius of the ball and a is the center.

Assume that B(E/2, a) is closed (since a similar open ball is contained in the closed one we can infer it is true as well). We can find balls around each x (contained in A) that can be described by B(E, x) (open or closed does not matter).

For B(E/2, a), we can always find d(x, a) <= E/2 < E...

I am having trouble describing the obvious assumption that the union of B(E, x)'s will contain (and more) all of B(E/2), do I need to use the triangle inequality? Or is a written explanation reason enough?

That doesn't make any sense at all. Are you sure you know what the definition of 'totally bounded' is? Isn't any bounded subset of R contained in an interval like [-M,M]?

 

[jdcasey9]

That doesn't make any sense at all. Are you sure you know what the definition of 'totally bounded' is? Isn't any bounded subset of R contained in an interval like [-M,M]?

For A subset in R I am trying to say that A is a subset of the union of balls that contain points in A (A=B(E/2, a) every other ball centered in A is B(E, x)). I have made the radiuses as such so that the Union of balls will always encompase every point in A. So, yes, every bounded subset of R is contained in an interval like [-M, M] (or (-M, M)), my bounded interval is the closed ball B(E/2, a)

The definition from notes for totally bounded:

(M,d) Fix E>0. A subset in M is an E-net if for all x is contained in M, there exists an a contained in A such that d(x, a)<E, equivalently, for all x contained in M, d(x,a)<E, equivalently, M is a subset of the Union of B(E,a). In words: "M is covered by E-balls".

 

[Dick]

For A subset in R I am trying to say that A is a subset of the union of balls that contain points in A (A=B(E/2, a) every other ball centered in A is B(E, x)). I have made the radiuses as such so that the Union of balls will always encompase every point in A. So, yes, every bounded subset of R is contained in an interval like [-M, M] (or (-M, M)), my bounded interval is the closed ball B(E/2, a)

The definition from notes for totally bounded:

(M,d) Fix E>0. A subset in M is an E-net if for all x is contained in M, there exists an a contained in A such that d(x, a)<E, equivalently, for all x contained in M, d(x,a)<E, equivalently, M is a subset of the Union of B(E,a). In words: "M is covered by E-balls".

That's the definition of E-net, not totally bounded. My definition says that M is totally bounded if for every e>0 there is a covering of M by a finite number of e-balls. Or, for every e>0 M contains a finite set A which is an e-net.

 

[jdcasey9]

Ok. I'm understanding the difference between the two, but I am not understanding how that changes my prospective proof.

 

[Dick]

Ok. I'm understanding the difference between the two, but I am not understanding how that changes my prospective proof.

I guess I just don't understand what your proof is trying to do. On the other hand it's pretty easy to show directly that the interval [-M,M] can be covered with a finite number of balls of radius e, no matter how small e is.

 

[jdcasey9]

Ok. So,

Fix E > 0. Let A = [-M, M] which is contained in R and let x=supA, y=infA. For any E, we know that x+E, y-E are outside of our interval and x-E, y+E are inside our interval. So to find the finitely many E-balls that cover [-M, M] would we just take any points M/2, -M/2 such that balls around them cover the entire interval and more? Like defining their radius to each be half of the interval?

 

[Dick]

Ok. So,

Fix E > 0. Let A = [-M, M] which is contained in R and let x=supA, y=infA. For any E, we know that x+E, y-E are outside of our interval and x-E, y+E are inside our interval. So to find the finitely many E-balls that cover [-M, M] would we just take any points M/2, -M/2 such that balls around them cover the entire interval and more? Like defining their radius to each be half of the interval?

You can't pick the radius of your balls, they have to be E. Suppose A is contained in [-M,M]. Then if you cover [-M,M] you cover A. To cover [-M,M] just pick the center of the balls to be -M, -M+E/2, -M+2*(E/2), -M+3*(E/2), ... How many do you need?

 

[jdcasey9]

We need enough so that the last one is centered around M. So if we continued in our pattern: -M, -M+E/2, -M+2*(E/2), -M+3*(E/2), ..., -M+2M would be last point, right? So we would need 4m/E points? (because x*(E/2) = 2M)

 

[Dick]

We need enough so that the last one is centered around M. So if we continued in our pattern: -M, -M+E/2, -M+2*(E/2), -M+3*(E/2), ..., -M+2M would be last point, right? So we would need 4m/E points? (because x*(E/2) = 2M)

More or less, yes. E might not divide evenly into M, but the exact count isn't important. What's important is that you can clearly cover any finite interval with a finite number of balls of size E.

 

[jdcasey9]

Ok, so to prove that a bounded subset of R is totally bounded:

Let E>0. Take A, a subset of [-M, M]. We can find finitely many balls such that their union covers [-M, M] by taking the balls B[E, -M], B[E, -M + E/2], B[E, -M + E], B[E, -M + 3E/2],..., B[E, M - E/2], B[E, M] (where E is the radius and the other is the center). Therefore, we have found a finite E-net that covers the interval [-M, M] and therefore covers A. So a bounded subset of R is totally bounded.

Is that sufficient enough?

 

[Dick]

Ok, so to prove that a bounded subset of R is totally bounded:

Let E>0. Take A, a subset of [-M, M]. We can find finitely many balls such that their union covers [-M, M] by taking the balls B[E, -M], B[E, -M + E/2], B[E, -M + E], B[E, -M + 3E/2],..., B[E, M - E/2], B[E, M] (where E is the radius and the other is the center). Therefore, we have found a finite E-net that covers the interval [-M, M] and therefore covers A. So a bounded subset of R is totally bounded.

Is that sufficient enough?

It expresses the idea well enough. If you want to make it a little neater you can also choose M to be a even multiple of E/2.

 

[jdcasey9]

Ok, but I just want to make sure that this is considered a sufficient proof for this problem. Nothing needs to be done more explicitly?